Question 155: Motion in 2D & Vectors

Question

A ball of mass m is thrown vertically upward. Another ball of mass $$2 \mathrm{~m}$$ is thrown at an angle $$	heta$$ with the vertical. Both the balls stay in air for the same period of time. The ratio of the heights attained by the two balls respectively is $$\frac{1}{x}$$. The value of x is _____________.

Accepted Answer

$$	herefore$$ $${u_1} = {u_2}\sin 	heta $$
$${{{H_1}} \over {{H_2}}} = {{{{u_1^2} \over {2g}}} \over {u_2^2{{{{\sin }^2}	heta } \over {2g}}}}$$
$$ = {\left( {{{{u_1}} \over {{u_2}\sin 	heta }}} ight)^2} = 1$$

Motion in 2D & Vectors

A ball of mass m is thrown vertically upward. Another ball of mass $2 \mathrm{~m} is thrown at an angle \theta with the vertical. Both the balls stay in air for the same period of time. The ratio of the heights attained by the two balls respectively is \frac{1}{x}$. The value of x is _____________.

Integer / Numeric Answer Type